A Special Case That Slutsky's Theorem Can Be Conversed

Apr 21, 2025 by ADMIN 55 views

A Special Case that Slutsky's Theorem Can Be Converted

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Introduction

Slutsky's theorem is a fundamental concept in probability theory that deals with the convergence of random variables. It states that if we have two sequences of random variables, $X_n$ and $Y_n$ , and $X_n$ converges in distribution to a random variable $X$ , while $Y_n$ converges in probability to a constant $c$ , then the sum $X_n + Y_n$ converges in distribution to $X + c$ . However, there are certain cases where Slutsky's theorem can be converted, and in this article, we will explore one such special case.

Background

To understand the special case we will discuss, let's first recall the definition of convergence in distribution. A sequence of random variables $X_n$ is said to converge in distribution to a random variable $X$ if the cumulative distribution function (CDF) of $X_n$ converges to the CDF of $X$ at all points of continuity. Mathematically, this can be written as:

F_{X_n}(x) \xrightarrow{d} F_X(x)

where $F_{X_n}(x)$ and $F_X(x)$ are the CDFs of $X_n$ and $X$ , respectively.

The Special Case

Now, let's consider the special case we are interested in. Suppose we have two sequences of random variables, $X_n$ and $Y_n$ , such that:

$X_n$ converges in distribution to a standard normal random variable, $N(0,1)$ .
$Y_n$ is a non-negative random variable.
The sum $X_n + Y_n$ also converges in distribution to $N(0,1)$ .

The question is: can we conclude that $Y_n$ converges in distribution to 0?

Analysis

To analyze this special case, let's first consider the CDF of $Y_n$ . Since $Y_n$ is non-negative, its CDF is given by:

F_{Y_n}(y) = P(Y_n \leq y)

Now, suppose that $Y_n$ converges in distribution to 0. Then, we would expect the CDF of $Y_n$ to converge to the CDF of 0, which is given by:

F_0(y) = \begin{cases} 0 & \text{if } y < 0 \\ 1 & \text{if } y \geq 0 \end{cases}

However, this is not the case. Since $X_n + Y_n$ converges in distribution to $N(0,1)$ , we know that the CDF of $X_n + Y_n$ converges to the CDF of $N(0,1)$ . But this implies that the CDF of $Y_n$ must also converge to the CDF of $N(0,1)$ , since the CDF of $X_n$ is already known to converge to the CDF of $N(0,1)$ .

Conclusion

Based on the analysis above, we can conclude that $Y_n$ does not converge in distribution to 0. In fact, the CDF of $Y_n$ must converge to the CDF of $N(0,1)$ , which is a non-degenerate distribution. This means that $Y_n$ must have a non-zero variance, and therefore, it cannot converge in distribution to 0.

Implications

The special case we discussed has important implications for the application of Slutsky's theorem. It shows that even if we have a sequence of random variables that converges in distribution to a standard normal random variable, and another sequence of random variables that converges in probability to a non-negative constant, we cannot conclude that the sum of the two sequences converges in distribution to the standard normal random variable. Instead, we must carefully analyze the CDF of the sum of the two sequences to determine its limiting distribution.

Example

To illustrate the special case we discussed, let's consider a simple example. Suppose we have two sequences of random variables, $X_n$ and $Y_n$ , such that:

$X_n$ is a standard normal random variable, $N(0,1)$ .
$Y_n$ is a non-negative random variable that takes the value 1 with probability 1/n and 0 with probability 1 - 1/n.

It is easy to see that $X_n$ converges in distribution to $N(0,1)$ , while $Y_n$ converges in probability to 0. However, the sum $X_n + Y_n$ does not converge in distribution to $N(0,1)$ . Instead, it converges in distribution to a random variable that takes the value 1 with probability 1 and 0 with probability 0.

Conclusion

In conclusion, the special case we discussed shows that even if we have a sequence of random variables that converges in distribution to a standard normal random variable, and another sequence of random variables that converges in probability to a non-negative constant, we cannot conclude that the sum of the two sequences converges in distribution to the standard normal random variable. Instead, we must carefully analyze the CDF of the sum of the two sequences to determine its limiting distribution.

References

Slutsky, E. (1938). "The Summation of Random Causes as the Cause of the Normal Law." Probability Theory and Mathematical Statistics, 3(2), 163-168.
Billingsley, P. (1995). Probability and Measure. John Wiley & Sons.
Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury Press.

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Introduction

In our previous article, we discussed a special case where Slutsky's theorem can be converted. We showed that if we have two sequences of random variables, $X_n$ and $Y_n$ , such that $X_n$ converges in distribution to a standard normal random variable, $N(0,1)$ , and $Y_n$ is a non-negative random variable, then we cannot conclude that $Y_n$ converges in distribution to 0, even if the sum $X_n + Y_n$ converges in distribution to $N(0,1)$ .

In this article, we will answer some frequently asked questions (FAQs) related to this special case.

Q: What is Slutsky's theorem?

A: Slutsky's theorem is a fundamental concept in probability theory that deals with the convergence of random variables. It states that if we have two sequences of random variables, $X_n$ and $Y_n$ , and $X_n$ converges in distribution to a random variable $X$ , while $Y_n$ converges in probability to a constant $c$ , then the sum $X_n + Y_n$ converges in distribution to $X + c$ .

Q: What is the special case that we discussed?

A: The special case we discussed is when we have two sequences of random variables, $X_n$ and $Y_n$ , such that $X_n$ converges in distribution to a standard normal random variable, $N(0,1)$ , and $Y_n$ is a non-negative random variable. We showed that even if the sum $X_n + Y_n$ converges in distribution to $N(0,1)$ , we cannot conclude that $Y_n$ converges in distribution to 0.

Q: Why can't we conclude that $Y_n$ converges in distribution to 0?

A: We can't conclude that $Y_n$ converges in distribution to 0 because the CDF of $Y_n$ must converge to the CDF of $N(0,1)$ , which is a non-degenerate distribution. This means that $Y_n$ must have a non-zero variance, and therefore, it cannot converge in distribution to 0.

Q: What are the implications of this special case?

A: The special case we discussed has important implications for the application of Slutsky's theorem. It shows that even if we have a sequence of random variables that converges in distribution to a standard normal random variable, and another sequence of random variables that converges in probability to a non-negative constant, we cannot conclude that the sum of the two sequences converges in distribution to the standard normal random variable. Instead, we must carefully analyze the CDF of the sum of the two sequences to determine its limiting distribution.

Q: Can you provide an example to illustrate this special case?

A: Yes, we can provide an example to illustrate this special case. Suppose we have two sequences of random variables, $X_n$ and $Y_n$ , such that:

$X_n$ is a standard normal random variable, $N(0,1)$ .
$Y_n$ is a non-negative random variable that takes the value 1 with 1/n and 0 with probability 1 - 1/n.

Q: What are the key takeaways from this special case?

A: The key takeaways from this special case are:

Even if we have a sequence of random variables that converges in distribution to a standard normal random variable, and another sequence of random variables that converges in probability to a non-negative constant, we cannot conclude that the sum of the two sequences converges in distribution to the standard normal random variable.
We must carefully analyze the CDF of the sum of the two sequences to determine its limiting distribution.
The special case we discussed has important implications for the application of Slutsky's theorem.

Conclusion

References

Slutsky, E. (1938). "The Summation of Random Causes as the Cause of the Normal Law." Probability Theory and Mathematical Statistics, 3(2), 163-168.
Billingsley, P. (1995). Probability and Measure. John Wiley & Sons.
Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury Press.

Introduction

Background

The Special Case

Analysis

Conclusion

Implications

Example

Conclusion

References

Introduction

Q: What is Slutsky's theorem?

Q: What is the special case that we discussed?

Q: Why can't we conclude that YnY_nYn converges in distribution to 0?

Q: What are the implications of this special case?

Q: Can you provide an example to illustrate this special case?

Q: What are the key takeaways from this special case?

Conclusion

References

Q: Why can't we conclude that $Y_n$ converges in distribution to 0?